NCERT Solutions for Class 10 Maths Chapter 11: Constructions

To construct: To construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3of the corresponding sides of the first triangle.

Steps of construction:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

Step 2

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

Step 3

Step 4

.

Step 5

Justification:

The construction can be justified by proving that

This justifies the construction.

 

To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are 7/5 of the corresponding sides of the first triangle.

Steps of construction:

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 3

Step 4

Step 5

.

Justification:

The construction can be justified by proving that

 

To construct: To construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then a triangle similar to it whose sides are 3/2 of the corresponding sides of the first triangle.

Steps of construction:

Step 1. Draw a line segment BC = 8 cm.

Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.

Step 3. With D as centre and radius 4 cm, draw an arc cutting XY at A.

Step 4. Join AB and AC. Here, ∆ABC is an isosceles whose base is 8 cm and altitude is 4 cm.

Step 5. Below BC, draw an acute angle ∠CBX.

Step 6. Along BX, mark three points

.

 

To construct: To construct a triangle ABC with side BC = 7 cm∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3of the corresponding sides of the first triangle ABC.

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

Steps of construction:

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 4 and 3)

Step  4

Step  5

Justification:

The construction can be justified by proving that

n ΔABC and ΔA'BC',

∠ABC = ∠A'BC' (Common)

∠ACB = ∠A'C'B (Corresponding angles)

∴ ΔABC ∼ ΔA'BC' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

This justifies the construction.

 

To construct: To construct a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm and then a triangle similar to it whose sides are 5/3 of the corresponding sides of the first triangle ABC.

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.

The required triangle can be drawn as follows.

Steps of construction:

Step 1

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

Step 2

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

Step 3

Draw a ray AX making an acute angle with AB, opposite to vertex C.

Step 4

Step 5

Step 6

Justification:

The construction can be justified by proving that

∠ABC = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

∴ ΔABC ∼ ΔAB'C' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

This justifies the construction.

Given: A line segment of length 7.6 cm.

To construct: To divide it in the ration 5: 8 and to measure the two parts.

Steps of construction:

Steps of Construction :

(i) Draw AB = 5.6 cm

(ii) At a draw an acute ∠BAX below base AB.

(iii) On AX make 5 + 8 i.e. 13 equal parts and mark them as A

On measuring, we get

AC = 3.1 cm,

CB = 4.5 cm,

Justification :

[Using basic proportionally theorem]

This shows that C divides AB in the ratio 5 : 8.

 

A ΔA'BC' whose sides areof the corresponding sides of ΔABC can be drawn as follows:

Steps of construction:

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4),

Step 4

Step 5

.

 

To construct: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.

Steps of Construction:

1. With ‘O’ as centre two circles are constructed with radii 4 cm and 6 cm.

2. Draw OM radius for small circle and drawn perpendicular at M which meets big circle at P and Q.

3. Now PQ is the tangnet drawn for small circle.

Tangent PMQ = 9 cm

Verification : In ⊥∆OMP, ∠M = 90°.

Similarly, MQ = 4.5 cm.

∴ PQ = PM + MQ = 4.5 + 4.5

∴ PQ = 9 cm.

 

 

Given: A circle whose centre is O and radius is 6 cm and a point P is 10 cm away from its centre.

To construct: To construct the pair of tangents to the circle and measure their lengths.

Steps of Construction:

(a) Join PO and bisect it. Let M be the mid-point of PO.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.

(c) Join PQ and PR.

Then PQ and PR are the required two tangents.

By measurement, PQ = PR = 8 cm

Justification:

Join OQ and OR.

∵ ∠OQP and ∠ORP are the angles in semi circles.

∠OQP = 90° = ∠ORP

Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at Q and R respectively.

 

To construct: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q.

Steps of Construction:

1. Draw a line segment PQ of 14 cm.

2. Now, mark the midpoint O of PQ.

3. Draw the perpendicular bisectors of PO and OQ which intersects at points R and S on PQ.

4. With centre R and radius RP draw a circle.

5. With centre S and radius, SQ draw a circle.

6. And now, with centre O and radius 3 cm draw another circle which intersects the previous circles at the points A, B, C, and D.

7. Finally, join PA, PB, QC and QD. Thus, PA, PB, QC, and QD are the required tangents.

 

To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600.

Steps of Construction:

Step I: Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm.

Step II: Produce OA to B such that OA = AB = 5 cm.

Step III: Taking A as the centre draw a circle of radius AO = AB = 5 cm.

Suppose it cuts the circle drawn in step I at P and Q.

Step IV: Join BP and BQ to get the desired tangents.

Justification: In OAP, we have

OA = OP = 5 cm (= Radius) Also,

AP = 5 cm (= Radius of circle with centre A)

∴ ∆OAP is equilateral.

⇒ ∠PAO = 60º ⇒ ∠BAP = 120º

In ∆BAP, we have

BA = AP and ∠BAP = 120º

∴ ∠ABP = ∠APB = 30º ⇒ ∠PBQ = 60º

 

To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.

Steps of Construction:

The tangents can be constructed on the given circles as follows

Step 1

Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.

Step 2

Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

Justification:

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠ASB = 90°

⇒ BS ⊥ AS

Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

 

To construct: A right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 900 .BD is the perpendicular from B on AC and the tangents from A to this circle.

Follow the given steps to construct the figure.

Steps of Construction:

Step 1

Draw a line BC of 8 cm length.

Step 2

Draw BX perpendicular to BC.

Step 3

Mark an arc at the distance of 6 cm on BX. Mark it as A.

Step 4

Join A and C. Thus, ∆ABC is the required triangle.

Step 5

With B as the centre, draw an arc on AC.

Step 6

Draw the bisector of this arc and join it with B. Thus, BD is perpendicular to AC.

Step 7

Now, draw the perpendicular bisector of BD and CD. Take the point of intersection as O.

Step 8

With O as the centre and OB as the radius, draw a circle passing through points B, C and D.

Step 9

Join A and O and bisect it. Let P be the midpoint of AO.

Step 10

Taking P as the centre and PO as its radius, draw a circle which will intersect the circle at point B and G. Join A and G.

Here, AB and AG are the required tangents to the circle from A.

Justification:

The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.

Then,

∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠AGE = 90°

⇒ EG ⊥ AG

Since EG is the radius of the circle, AG has to be a tangent of the circle.

Already, ∠B = 90°

⇒ AB ⊥ BE

Since BE is the radius of the circle, AB has to be a tangent of the circle.

 

To construct: A circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Steps of Construction:

The required tangents can be constructed on the given circle as follows.

Step 1

Draw a circle with the help of a bangle.

Step 2

Take a point P outside this circle and take two chords QR and ST.

Step 3

Draw perpendicular bisectors of these chords. Let them intersect each other at point O.

Step 4

Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre, draw a circle of radius OU, which will intersect the circle at V and W. Join PV and PW.

PV and PW are the required tangents.

Justification:

The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW

 

Then,

We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠PVO = 90°

⇒ OV ⊥ PV

Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.